2. Not all languages are regular
So what happens to the languages
which are not regular?
Can we still come up with a language
recognizer?
i.e., something that will accept (or reject)
strings that belong (or do not belong) to the
language?
2
3. 3
Context-Free Languages
A language class larger than the class of regular
languages
Supports natural, recursive notation called “context-
free grammar”
Applications:
Parse trees, compilers
XML
Regular
(FA/RE)
Context-
free
(PDA/CFG)
4. 4
An Example
A palindrome is a word that reads identical from both
ends
E.g., madam, redivider, malayalam, 010010010
Let L = { w | w is a binary palindrome}
Is L regular?
No.
Proof:
Let w=0N10N (assuming N to be the p/l constant)
By Pumping lemma, w can be rewritten as xyz, such that xykz is also L
(for any k≥0)
But |xy|≤N and y≠
==> y=0+
==> xykz will NOT be in L for k=0
==> Contradiction
5. 5
But the language of
palindromes…
is a CFL, because it supports recursive
substitution (in the form of a CFG)
This is because we can construct a
“grammar” like this:
1. A ==>
2. A ==> 0
3. A ==> 1
4. A ==> 0A0
5. A ==> 1A1
Terminal
Productions
Variable or non-terminal
How does this grammar work?
Same as:
A => 0A0 | 1A1 | 0 | 1 |
6. How does the CFG for
palindromes work?
An input string belongs to the language (i.e.,
accepted) iff it can be generated by the CFG
Example: w=01110
G can generate w as follows:
1. A => 0A0
2. => 01A10
3. => 01110
6
G:
A => 0A0 | 1A1 | 0 | 1 |
Generating a string from a grammar:
1. Pick and choose a sequence
of productions that would
allow us to generate the
string.
2. At every step, substitute one variable
with one of its productions.
7. 7
Context-Free Grammar:
Definition
A context-free grammar G=(V,T,P,S), where:
V: set of variables or non-terminals
T: set of terminals (= alphabet U {})
P: set of productions, each of which is of the form
V ==> 1 | 2 | …
Where each i is an arbitrary string of variables and
terminals
S ==> start variable
CFG for the language of binary palindromes:
G=({A},{0,1},P,A)
P: A ==> 0 A 0 | 1 A 1 | 0 | 1 |
8. 8
More examples
Parenthesis matching in code
Syntax checking
In scenarios where there is a general need
for:
Matching a symbol with another symbol, or
Matching a count of one symbol with that of
another symbol, or
Recursively substituting one symbol with a string
of other symbols
9. 9
Example #2
Language of balanced paranthesis
e.g., ()(((())))((()))….
CFG?
G:
S => (S) | SS |
How would you “interpret” the string “(((()))()())” using this grammar?
10. 10
Example #3
A grammar for L = {0m1n | m≥n}
CFG? G:
S => 0S1 | A
A => 0A |
How would you interpret the string “00000111”
using this grammar?
11. 11
Example #4
A program containing if-then(-else) statements
if Condition then Statement else Statement
(Or)
if Condition then Statement
CFG?
12. More examples
L1 = {0n | n≥0 }
L2 = {0n | n≥1 }
L3={0i1j2k | i=j or j=k, where i,j,k≥0}
L4={0i1j2k | i=j or i=k, where i,j,k≥1}
12
13. 13
Applications of CFLs & CFGs
Compilers use parsers for syntactic checking
Parsers can be expressed as CFGs
1. Balancing paranthesis:
B ==> BB | (B) | Statement
Statement ==> …
2. If-then-else:
S ==> SS | if Condition then Statement else Statement | if Condition
then Statement | Statement
Condition ==> …
Statement ==> …
3. C paranthesis matching { … }
4. Pascal begin-end matching
5. YACC (Yet Another Compiler-Compiler)
14. 14
More applications
Markup languages
Nested Tag Matching
HTML
<html> …<p> … <a href=…> … </a> </p> … </html>
XML
<PC> … <MODEL> … </MODEL> .. <RAM> …
</RAM> … </PC>
15. 15
Tag-Markup Languages
Roll ==> <ROLL> Class Students </ROLL>
Class ==> <CLASS> Text </CLASS>
Text ==> Char Text | Char
Char ==> a | b | … | z | A | B | .. | Z
Students ==> Student Students |
Student ==> <STUD> Text </STUD>
Here, the left hand side of each production denotes one non-terminals
(e.g., “Roll”, “Class”, etc.)
Those symbols on the right hand side for which no productions (i.e.,
substitutions) are defined are terminals (e.g., ‘a’, ‘b’, ‘|’, ‘<‘, ‘>’, “ROLL”,
etc.)
16. 16
Structure of a production
A =======> 1 | 2 | … | k
head body
derivation
1. A ==> 1
2. A ==> 2
3. A ==> 3
…
K. A ==> k
The above is same as:
17. 17
CFG conventions
Terminal symbols <== a, b, c…
Non-terminal symbols <== A,B,C, …
Terminal or non-terminal symbols <== X,Y,Z
Terminal strings <== w, x, y, z
Arbitrary strings of terminals and non-
terminals <== , , , ..
18. 18
Syntactic Expressions in
Programming Languages
result = a*b + score + 10 * distance + c
Regular languages have only terminals
Reg expression = [a-z][a-z0-1]*
If we allow only letters a & b, and 0 & 1 for
constants (for simplification)
Regular expression = (a+b)(a+b+0+1)*
terminals variables Operators are also
terminals
19. 19
String membership
How to say if a string belong to the language
defined by a CFG?
1. Derivation
Head to body
2. Recursive inference
Body to head
Example:
w = 01110
Is w a palindrome?
Both are equivalent forms
G:
A => 0A0 | 1A1 | 0 | 1 |
A => 0A0
=> 01A10
=> 01110
20. 20
Simple Expressions…
We can write a CFG for accepting simple
expressions
G = (V,T,P,S)
V = {E,F}
T = {0,1,a,b,+,*,(,)}
S = {E}
P:
E ==> E+E | E*E | (E) | F
F ==> aF | bF | 0F | 1F | a | b | 0 | 1
21. 21
Generalization of derivation
Derivation is head ==> body
A==>X (A derives X in a single step)
A ==>*G X (A derives X in a multiple steps)
Transitivity:
IFA ==>*GB, and B ==>*GC, THEN A ==>*G C
22. 22
Context-Free Language
The language of a CFG, G=(V,T,P,S),
denoted by L(G), is the set of terminal
strings that have a derivation from the
start variable S.
L(G) = { w in T* | S ==>*G w }
23. 23
Left-most & Right-most
Derivation Styles
Derive the string a*(ab+10) from G:
E
==> E * E
==> F * E
==> aF * E
==> a * E
==> a * (E)
==> a * (E + E)
==> a * (F + E)
==> a * (aF + E)
==> a * (abF + E)
==> a * (ab + E)
==> a * (ab + F)
==> a * (ab + 1F)
==> a * (ab + 10F)
==> a * (ab + 10)
E =*=>G a*(ab+10)
Left-most
derivation:
E
==> E * E
==> E * (E)
==> E * (E + E)
==> E * (E + F)
==> E * (E + 1F)
==> E * (E + 10F)
==> E * (E + 10)
==> E * (F + 10)
==> E * (aF + 10)
==> E * (abF + 0)
==> E * (ab + 10)
==> F * (ab + 10)
==> aF * (ab + 10)
==> a * (ab + 10)
Right-most
derivation:
G:
E => E+E | E*E | (E) | F
F => aF | bF | 0F | 1F |
Always
substitute
leftmost
variable
Always
substitute
rightmost
variable
24. 24
Leftmost vs. Rightmost
derivations
Q1) For every leftmost derivation, there is a rightmost
derivation, and vice versa. True or False?
Q2) Does every word generated by a CFG have a
leftmost and a rightmost derivation?
Q3) Could there be words which have more than one
leftmost (or rightmost) derivation?
True - will use parse trees to prove this
Yes – easy to prove (reverse direction)
Yes – depending on the grammar
25. How to prove that your CFGs
are correct?
(using induction)
25
26. 26
CFG & CFL
Theorem: A string w in (0+1)* is in
L(Gpal), if and only if, w is a palindrome.
Proof:
Use induction
on string length for the IF part
On length of derivation for the ONLY IF part
Gpal:
A => 0A0 | 1A1 | 0 | 1 |
28. 28
Parse Trees
Each CFG can be represented using a parse tree:
Each internal node is labeled by a variable in V
Each leaf is terminal symbol
For a production, A==>X1X2…Xk, then any internal node
labeled A has k children which are labeled from X1,X2,…Xk
from left to right
A
X1 Xi Xk
… …
Parse tree for production and all other subsequent productions:
A ==> X1..Xi..Xk
29. 29
Examples
E
E + E
F F
a 1
Parse tree for a + 1
A
0 A 0
1 1
A
Parse tree for 0110
Recursive
inference
Derivation
G:
E => E+E | E*E | (E) | F
F => aF | bF | 0F | 1F | 0 | 1 | a | b
G:
A => 0A0 | 1A1 | 0 | 1 |
30. 30
Parse Trees, Derivations, and
Recursive Inferences
A
X1 Xi Xk
… …
Recursive
inference
Derivation
Production:
A ==> X1..Xi..Xk
Parse tree
Left-most
derivation
Right-most
derivation
Derivation
Recursive
inference
31. 31
Interchangeability of different
CFG representations
Parse tree ==> left-most derivation
DFS left to right
Parse tree ==> right-most derivation
DFS right to left
==> left-most derivation == right-most
derivation
Derivation ==> Recursive inference
Reverse the order of productions
Recursive inference ==> Parse trees
bottom-up traversal of parse tree
33. 33
CFLs & Regular Languages
A CFG is said to be right-linear if all the
productions are one of the following two
forms: A ==> wB (or) A ==> w
Theorem 1: Every right-linear CFG generates
a regular language
Theorem 2: Every regular language has a
right-linear grammar
Theorem 3: Left-linear CFGs also represent
RLs
Where:
• A & B are variables,
• w is a string of terminals
What kind of grammars result for regular languages?
34. Some Examples
34
A B C
0
1 0
1 0,1
A B C
0
1 0
1
1
0
A => 01B | C
B => 11B | 0C | 1A
C => 1A | 0 | 1
Right linear CFG? Right linear CFG? Finite Automaton?
36. 36
Ambiguity in CFGs
A CFG is said to be ambiguous if there
exists a string which has more than one
left-most derivation
LM derivation #1:
S => AS
=> 0A1S
=>0A11S
=> 00111S
=> 00111
Example:
S ==> AS |
A ==> A1 | 0A1 | 01
Input string: 00111
Can be derived in two ways
LM derivation #2:
S => AS
=> A1S
=> 0A11S
=> 00111S
=> 00111
37. 37
Why does ambiguity matter?
string = a * b + c
E ==> E + E | E * E | (E) | a | b | c | 0 | 1
• LM derivation #1:
•E => E + E => E * E + E
==>* a * b + c
• LM derivation #2
•E => E * E => a * E =>
a * E + E ==>* a * b + c
E
E + E
E * E
a b
c
(a*b)+c
E
E * E
E
+
E
a
b c
a*(b+c)
Values are
different !!!
The calculated value depends on which
of the two parse trees is actually used.
38. 38
Removing Ambiguity in
Expression Evaluations
It MAY be possible to remove ambiguity for
some CFLs
E.g.,, in a CFG for expression evaluation by
imposing rules & restrictions such as precedence
This would imply rewrite of the grammar
Precedence: (), * , +
How will this avoid ambiguity?
E => E + T | T
T => T * F | F
F => I | (E)
I => a | b | c | 0 | 1
Modified unambiguous version:
Ambiguous version:
E ==> E + E | E * E | (E) | a | b | c | 0 | 1
39. 39
Inherently Ambiguous CFLs
However, for some languages, it may not be
possible to remove ambiguity
A CFL is said to be inherently ambiguous if
every CFG that describes it is ambiguous
Example:
L = { anbncmdm | n,m≥ 1} U {anbmcmdn | n,m≥ 1}
L is inherently ambiguous
Why?
Input string: anbncndn
